exponential dichotomies for linear non-autonomous functional diverential equations of...

Exponential Dichotomies forLinear Non-autonomous

Functional Differential Equations of Mixed TypeJORG HARTERICH, BJORN SANDSTEDE, AND ARND SCHEEL

ABSTRACT. Functional differential equations with forward andbackward delays arise naturally, for instance, in the study of trav-elling waves in lattice equations and as semi-discretizations ofpartial differential equations (PDEs) on unbounded domains.Linear functional differential equations of mixed type are typi-cally ill-posed, i.e., there exists, in general, no solution to a giveninitial condition. We prove that Fredholm properties of theseequations imply the existence of exponential dichotomies. Expo-nential dichotomies can be used in discretized PDEs and in lat-tice differential equations to construct multi-pulses, to performEvans-function type calculations, and to justify numerical com-putations using artificial boundary conditions.

1. INTRODUCTION

We are interested in linear non-autonomous functional differential equations

(1.1) v′(ξ) =m∑

j=−mAj(ξ)v(ξ + j), ξ ∈ R

of mixed type, where v ∈ Cn, and Aj(ξ) are continuous functions with valuesin Cn×n for |j| ≤ m. Before we motivate our interest in this equation and lista number of applications that we have in mind, we discuss a few properties of(1.1). The most important feature of (1.1), at least for the purpose of this paper, isthat the associated initial-value problem is ill-posed. To make this statement moreprecise, we should first explain in what sense we want to solve (1.1): We say thata function v(ξ) satisfies (1.1) on an interval J = [a, b], where a = ∞ and b = ∞

1081Indiana University Mathematics Journal c©, Vol. 51, No. 5 (2002)

1082 JORG HARTERICH. BJORN SANDSTEDE, AND ARND SCHEEL

are allowed, if v ∈ L2loc([a−m, b+m],Cn) ∩H1

loc([a, b],Cn), and (1.1) is metin L2

loc([a, b],Cn). The initial-value problem associated with (1.1) is given by

(1.2) v′(ξ) =m∑

j=−mAj(ξ)v(ξ + j), v|[−m,m] = φ,

where φ is a given function defined on [−m,m]. Unfortunately, for a givenfunction φ, there is in general no solution, in the above sense, to (1.2) on anynontrivial interval that contains ξ = 0. A simple counterexample (see [15]) isprovided by the equation

(1.3) v′(ξ) = v(ξ − 1)+ v(ξ + 1), v|[−m,m] = 1

with v ∈ C. The only function that could possibly be a solution of this initial-value problem is given by v(ξ) = (−1)k for ξ ∈ (2k−1, 2k+1] with k ∈ N;this function, however, is not even continuous. Seeking solutions of the formv(ξ) = eλξ , we see that the characteristic eigenvalue equation associated with(1.3) is

λ = e−λ + eλ.

This equation has solutions λ ∈ C with Reλ arbitrarily large and also admitssolutions for which −Reλ is arbitrarily large. Therefore, the linear equation (1.3)does not generate a semiflow on any space that contains all its eigenfunctions. Thisexplains why the initial-value problem (1.2) associated with (1.1) is ill-posed. Infact, functional differential equations of mixed type behave quite similar to ellipticPDEs when considered as initial-value problems. Note also that solving (1.1)forward or backward in the time variable ξ is equally difficult.

Since we cannot solve (1.2) for all φ, we should therefore find those functionsφ for which a solution to (1.2) exists on either R+ or R−. In particular, we wouldexpect to be able to solve the linear autonomous equation (1.3) for ξ > 0 forany initial condition φ that is a superposition of eigenfunctions associated withstable eigenvalues (i.e., eigenvalues with negative real part). In fact, the resultingsolution should decay to zero exponentially as ξ → ∞. Analogously, we shouldbe able to solve (1.3) on R− for any initial condition φ that is a superpositionof eigenfunctions associated with unstable eigenvalues (i.e., eigenvalues of posi-tive real part), and the solution should decay exponentially as ξ → −∞. Usingresults from [1] about the characteristic equation, Rustichini [15] proved theseassertions for autonomous equations. His result leads naturally to the questionhow large the closure of all eigenfunctions associated with either stable or unstableeigenvalues is. Indeed, the sum of the resulting closed spaces gives the functionspace on which we can construct solutions to (1.1) on either R+ or R−. The dif-ficulty in determining whether this sum is the entire underlying function space,i.e., whether the set of eigenfunctions is complete, lies in the problem of excludingsolutions that decay super-exponentially, so-called small solutions. Verduyn Lunel

Linear Non-autonomous Functional Differential Equations 1083

[19] gave conditions that guarantee that the set of eigenfunctions associated withan autonomous functional differential equation is complete.

In this paper, we address the above issues for non-autonomous functionaldifferential equations of mixed type. The obvious difficulty is that the spaces onwhich we can solve (1.1) forward or backward in time will depend on ξ. It is notapriori clear which spaces will replace the unstable and stable eigenspaces that wereso useful for autonomous equations. It turns out that the correct notion in thenon-autonomous setup are exponential dichotomies. An exponential dichotomyformalizes the idea of solving (1.1) either forward or backward in ξ for initialconditions in certain complementary subspaces even though these subspaces willdepend on ξ.

To formulate the definition of exponential dichotomies in the present context,it is convenient to introduce the following notation, which we shall use frequently.For a given function v : [−m,∞) → Cn, we define vξ : [−m,m] → Cn viavξ(η) := v(ξ + η) for η ∈ [−m,m].

Definition 1.1 ([14]). Let J = [a, b] be R+, R− or R. Equation (1.1)is said to have an exponential dichotomy on the interval J if there exist posi-tive constants K and κ, and a strongly continuous family of projections P(ξ) :L2([−m,m],Cn) → L2([−m,m],Cn) such that the following is true for anyφ ∈ L2([−m,m],Cn) and ζ ∈ J.

(i) There exists a unique solution v on [ζ, b] of (1.1) such that vζ = P(ζ)φ.In addition, vξ ∈ R(P(ξ)) and

‖vξ‖L2([−m,m],Cn) ≤ Ke−κ|ξ−ζ|‖φ‖L2([−m,m],Cn)

for all ξ ≥ ζ with ξ, ζ ∈ J.(ii) There exists a unique solution v on [a, ζ] of (1.1) such that vζ =

(id−P(ζ))φ. In addition, vξ ∈ N(P(ξ)) and

‖vξ‖L2([−m,m],Cn) ≤ Ke−κ|ξ−ζ|‖φ‖L2([−m,m],Cn)

for all ξ ≤ ζ with ξ, ζ ∈ J.

Exponential dichotomies have been shown to exist in ordinary differentialequations [4], parabolic PDEs [7] and delay equations [6], where the unstablesubspace N(P(ξ)) is always finite-dimensional, and the initial-value problem iswell-posed. In [14], the existence of exponential dichotomies has been establishedfor elliptic PDEs on unbounded domains. Here, both R(P(ξ)) and N(P(ξ)) areinfinite-dimensional, and the initial-value problem is ill-posed.

Associated with (1.1) is the operator

L : H1(R,Cn) -→ L2(R,Cn), (Lv)(ξ) = dvdξ(ξ)−

m∑j=−m

Aj(ξ)v(ξ + j)


and its formal adjoint L∗, defined on the same spaces, given by

(L∗w)(ξ) = −dwdξ(ξ)−

m∑j=−m

A∗j (ξ − j)w(ξ − j).

We need the following weak uniqueness assumption.

Hypothesis 1.1. If v is in the null space of L or the adjoint operator L∗ suchthat v|[ξ0−m,ξ0+m] = 0 for some ξ0, then v vanishes identically.

Our main result is the following theorem.

Theorem 1.1. If L is a Fredholm operator and if Hypothesis 1.1 is met, then(1.1) has exponential dichotomies on R+ and on R−.

A weaker but perhaps more explicit version of the above theorem is the fol-lowing statement.

Theorem 1.2. If (1.1) is asymptotically hyperbolic (see Definition 2.1 below)and if neither detAm(ξ) nor detA−m(ξ) vanish on any open interval, then (1.1)has exponential dichotomies on R+ and on R−.

Analogous results have been shown, independently and simultaneously, in[12].

The existence of exponential dichotomies on R+ and R− has a number ofconsequences: for instance, the null space and the orthogonal complement of therange of L are isomorphic to the spaces R(P+(0)) ∩ N(P−(0)) and (R(P+(0)) +N(P−(0)))⊥, respectively. In addition, it is possible to characterize the Fredholmindex of L by the difference of relative Morse indices. We refer to [18] for details.

Lastly, we motivate why functional differential equations are interesting andoutline some applications of exponential dichotomies that we intend to pursue infuture work. Linear non-autonomous functional differential equations of mixedtype arise in many different problems. We may, for instance, be interested intravelling waves of lattice differential equations

∂tuk =m∑

j=−mfj(uk+j), k ∈ Z,

where uj = uj(t) for j ∈ Z. A travelling-wave solution is a function ϕ(ξ) suchthat, for some wave speed c ∈ R, we have uk(t) = ϕ(k + ct) for t ∈ R andk ∈ Z. Upon substituting this expression for uk into the above lattice equation,we obtain

cϕ′(ξ) =m∑

j=−mfj(ϕ(ξ + j)), ξ ∈ R,


where we set ξ = k+ ct. The linearization about the wave ϕ(ξ) is then given by

cv′(ξ) =m∑

j=−mDfj(ϕ(ξ + j))v(ξ + j) =:

m∑j=−m

Aj(ξ)v(ξ + j)

which is of the form (1.1) provided the wave speed c does not vanish. Note that,if c = 0, then the above equation is a difference equation. A second example aresemi-discretizations of parabolic PDEs such as

∂tu = D∂2xu+ f(u), u ∈ Rn, x ∈ R

that admit travelling-wave solutions which connect two, possibly different, homo-geneous equilibria. Since such equations are often too complicated to allow for acomplete analysis, numerical methods have to be employed to compute travellingwaves and to continue them in parameter space. An important question is thento which extent the numerical scheme is able to reproduce travelling waves of theoriginal PDE and whether the stability properties of the wave are retained upondiscretizing. We shall investigate these issues in a simplified setting: instead ofconsidering a fully discrete numerical scheme, we may study semi-discretizations,i.e., equations where only the spatial derivatives are replaced by finite differenceapproximations. The resulting lattice equations are of the form

∂tu(x, t) =m∑

j=−mαju(x+jh, t)+ f(u(x, t)),

where the coefficients αj may depend on the mesh size h. Let ξ := (x + ct)/h,then a travelling wave of the form u(x, t) =ϕ((x+ct)/h) satisfies the nonlinearfunctional differential equation

chϕ′(ξ) =

m∑j=−m

αjϕ(ξ + j)+ f(ϕ(ξ)).

We assume that we have found a solution ϕ(ξ) of this equation and consider thelinearization

chv′(ξ) =

m∑j=−m

αjv(ξ + j)+Df(ϕ(ξ))v(ξ)

about the wave. If the wave speed c ≠ 0 is not zero, we obtain a functional dif-ferential equation of mixed type as in our first example. Exponential dichotomiesprovide a useful tool to investigate such equations. In a nutshell (see [18] fora more comprehensive discussion), exponential dichotomies allow for a much


more refined perturbation analysis compared with, for instance, Fredholm prop-erties. One example where dichotomies are useful is in providing correct choicesof boundary conditions so that (1.1) truncated to an interval (−L, L) with L� 1is well-posed (see [10] and the references therein). Dichotomies are also useful inthe construction of Evans functions that can be used to investigate linear stabilityof travelling waves (see, for instance, [2, 18, 17] and references therein). Lastly,exponential dichotomies can be used to construct new patterns, such as periodicor multi-hump waves, from a given travelling wave by using, for example, Lin’smethod [9, 16]. Some of the above issues will be investigated in more detail in aforthcoming article.

This paper is organized as follows. In Section 2, we formulate (1.1) as anevolution problem and introduce several operators relevant to that formulation.In Section 3, we then show the existence of exponential dichotomies for the cor-responding autonomous equation. The analysis is similar to the one given in[15] and uses additional results from [19]. Lastly, in Section 4, we consider thenon-autonomous equation for which we prove the existence of exponential di-chotomies following the strategy in [18].Acknowledgments. J. Harterich was supported by the Deutsche Forschungs-gemeinschaft under grant Ha 3008/1-1. B. Sandstede was partially supported bythe National Science Foundation under grant DMS-9971703 and by an AlfredP. Sloan Research Fellowship.

2. THE OPERATORS L AND TIn this section we describe two linear operators that can be associated with thelinear functional differential equation

(2.1) v′(ξ) =m∑

j=−mAj(ξ)v(ξ + j)

and show how they are related.

2.1. The operators L and L∗. We define the closed and densely definedlinear operator

L : D(L) = H1(R,Cn) -→ L2(R,Cn) ,

(Lv)(ξ) = dvdξ(ξ)−

m∑j=−m

Aj(ξ)v(ξ + j)

associated with equation (2.1). The adjoint operator L∗ of L is given by

(L∗w)(ξ) = −dwdξ(ξ)−

m∑j=−m

A∗j (ξ − j)w(ξ − j).


It is easily checked that∫∞−∞(Lv)(ξ) ·w(ξ)dξ =

∫∞−∞v(ξ) · (L∗w)(ξ)dξ

for all v,w ∈ H1(R,Cn), where we denote the scalar product in Cn by a · b =∑nk=1 akbk for a,b ∈ Cn. The operator L has particularly nice properties if the

coefficients Aj satisfy a certain hyperbolicity condition.

Definition 2.1. The linear functional differential equation (2.1) is calledasymptotically constant if the limits A±j := limξ→±∞Aj(ξ) exist for all j. Theequation is called asymptotically hyperbolic if it is asymptotically constant and ifthe characteristic equations

det∆±(µ) := det( m∑j=−m

A±j eµj − µ id)= 0

associated with the limiting equations at ξ = ±∞ have no solutions µ on theimaginary axis. If the coefficients do not depend on ξ, then we call (2.1) hyperbolicif det∆(µ) has no purely imaginary zeros µ.

The following result is due to Mallet-Paret [11].

Proposition 2.1 ([11]). If L is asymptotically hyperbolic, then L is a Fredholmoperator.

2.2. The operator T . A different way of viewing (2.1) is to write it in theform

(2.2)dVdξ(ξ) =A(ξ)V(ξ)

where, for each fixed ξ ∈ R, we define

A(ξ) : D(A(ξ)) = Y 1 -→ Y ,(φa

)7 -→

dφdη

A0(ξ)a+∑

1≤|j|≤mAj(ξ)φ(j)

with

Y := L2([−m,m],Cn)× Cn

Y 1 := {(φ,a) ∈ H1([−m,m],Cn)×Cn; φ(0) = a}.(2.3)

The space Y 1 is well defined since H1([−m,m],Cn) is embedded inC0([−m,m],Cn). Note also that Y 1 is dense in Y since the set of C1-functions


φ with φ(0) = a is dense in the set of step functions in the L2-norm. The op-erator A(ξ) has domain Y 1 and is closed for any fixed ξ. In the case of constantcoefficients, we write A0 instead of A(ξ).

Before we define the operator T , we state the following lemma that we usebelow to define the domain of T .

Lemma 2.1. Using the notation I = [−m,m] and (ξ, η) ∈ R×I, we have thatL2(R, L2(I,Cn)) = L2(R×I,Cn). Furthermore, there is a constant C with the follow-ing property. If φ ∈ L2(R × I,Cn) such that the weak derivative(∂ξ − ∂η)φ ∈ L2(R× I,Cn) exists, then φ(·, k) ∈ L2(R,Cn) for every fixed k ∈ Iand φ(0, ·) ∈ L2(I,Cn); in addition, we have

‖φ(0, ·)‖L2(I,Cn)+‖φ(·, k)‖L2(R,Cn) ≤ C(‖φ‖L2(R×I,Cn)+‖(∂ξ−∂η)φ‖L2(R×I,Cn)).

Proof. The identity L2(R, L2(I,Cn)) = L2(R × I,Cn) is a consequence ofFubini’s theorem. Upon introducing the coordinates (ξ, η) = (ξ+η, η) ∈ R× I,we see that φ(ξ,η) ∈ L2(R × I,Cn) with (∂ξ − ∂η)φ(ξ, η) ∈ L2(R × I,Cn)if, and only if, φ(ξ, η) := φ(ξ−η, η) ∈ L2(R,H1(I,Cn)). In particular, φ ∈L2(R, C0(I,Cn)), and we conclude that φ(·, k) ∈ L2(R,Cn) for every fixed k ∈ I.Hence, φ(ξ, k) = φ(ξ+k, k) ∈ L2(R,Cn) for fixed k. Lastly, for any such φ, wealso have thatφ(0, η) = φ(η, η) exists for almost every η ∈ I. It is straightforwardto see that the L2-norm of φ(η, η) can be bounded by the L2(R× I)-norms of φand (∂ξ − ∂η)φ upon using the coordinates (ξ, η). ❐

Associated with (2.2) is the operator

T : L2(R, Y ) -→ L2(R, Y ), V 7 -→ dVdξ(·)−A(·)V(·)

(φ,a) 7 -→dφ

dξ− dφ

dη,dadξ(ξ)−A0(ξ)a(ξ)−

∑1≤|j|≤m

Aj(ξ)[φ(ξ)](j)

which is considered as an unbounded operator on L2(R, Y ) with domain

(2.4) D(T ) = {(φ,a) ∈ L2(R, Y ); (∂ξ − ∂η)φ ∈ L2(R× I,Cn),a ∈ H1(R,Cn), [φ(ξ)](0) = a(ξ) ∀ξ}

where we use the notation I = [−m,m]. Note that D(T ) is well-defined owingto Lemma 2.1. Using again Lemma 2.1, it is not difficult to prove that T is closedand densely defined.

Note that we have L2(R, Y 1) ∩ H1(R, Y ) ⊂ D(T ). It is tempting to takeL2(R, Y 1) ∩ H1(R, Y ) as the domain of T . The operator T is, however, notclosed if considered with this domain.


The following lemma shows how the null spaces of the operators L and T arerelated.

Lemma 2.2. If a function v ∈ H1(R,Cn) satisfies Lv = 0, then V(ξ) :=(v|[ξ−m,ξ+m], v(ξ)) satisfies V ∈ D(T ) and T V = 0. Conversely, if V = (φ,a) ∈D(T ) satisfies T V = 0, then a ∈ H1(R,Cn) satisfies La = 0. In particular,N(L) � N(T ).

Proof. If v ∈ H1(R,Cn) is a solution to Lv = 0, then v is in fact of classC∞. To see this, it suffices to show that v ∈ Ck([−`, `],Cn) for arbitrarily largenumbers k and `. By the Sobolev Embedding Theorem, v is in C0 on the in-terval [−`−mk, `+mk]. Since v satisfies (2.1), it is C1 on the smaller interval[−`−(k− 1)m, `+(k−1)m]. Inductively one can then show that v is indeed ofclass Ck on [−`, `]. As a consequence, V(ξ) := (vξ, v(ξ)) is a classical solutionof (2.2). It remains to check that V is in the domain of T . Since v ∈ H1(R,Cn),we know that

‖V‖L2(R,Y 1) =∫R

∫m−m(|vξ(η)|2 + |∂ηvξ(η)|2)dηdξ + ‖v‖H1(R,Cn)

=∫R

∫m−m(|v(ξ + η)|2 + |v′(ξ + η)|2)dηdξ + ‖v‖H1(R,Cn)

= (2m+ 1)‖v‖H1(R,Cn),

and we conclude that V is in L2(R, Y 1). Note that v ∈ H1(R,Cn) is in the do-main of the derivative which is the generator of the shift semigroup on L2(R,Cn).Therefore,

ddξv(ξ + ·) = d

dηv(ξ + ·).

Hence V is in H1(R, Y ) and therefore indeed in L2(R, Y 1)×H1(R, Y ) ⊂ D(T ).To show the other direction, assume that V = (φ,a) ∈ D(T ) satisfies T V =

0. It follows from the definition of D(T ) that v(ξ) = a(ξ) = [φ(ξ)](0) iswell-defined and v ∈ H1(R,Cn). It remains to show that a|[ξ−m,ξ+m] = φ(ξ),i.e., [φ(ξ + η)](0) = [φ(ξ)](η). As in Lemma 2.1, we use the coordinates(ξ, η) = (ξ+η, η) ∈ R×I and set φ(ξ, η) := [φ(ξ− η)](η). From d

dξφ =ddηφ,

we conclude that φ(ξ, η) = φ(ξ,0) for almost every ξ, and therefore for every ξas [φ(ξ)](0) is continuous. Hence, [φ(ξ − η)](η) = [φ(ξ)](0) for every ξ andη, and we conclude that [φ(ξ + η)](0) = [φ(ξ)](η). ❐

2.3. The adjoint operator T ∗. We denote by 〈·, ·〉 the inner product⟨(φa

),(ψb

)⟩:=∫∞−∞

∫m−mφ(ξ,η) ·ψ(ξ,η)dηdξ +

∫∞−∞a(ξ) · b(ξ)dξ

on L2(R, Y ).


Remark 2.1. Let j be an integer with −m ≤ j < m. If ψ ∈L2(R × [j, j+1],Cn) with weak derivative (∂ξ − ∂η)ψ ∈ L2(R × (j, j+1),Cn),thenψ(·, j) and ψ(·, j+1) are in L2(R,Cn) by Lemma 2.1. We use the notationψ(·, j+) := ψ(·, j) and ψ(·, (j+1)−) := ψ(·, j+1). For convenience, we alsodefine ψ(·,m+) = ψ(·,−m−) = 0.

Lemma 2.3. The adjoint operator T ∗ is given by

T ∗ : L2(R, Y ) -→ L2(R, Y ),

(ψ,b) 7 -→(−dψ

dξ+ dψ

dη,−db

dξ−A∗0 (ξ)b +ψ(·,0−)−ψ(·,0+)

)

with

D(T ∗) = {(ψ,b) ∈ L2(R, Y ); (∂ξ − ∂η)ψ ∈ L2(R× (j, j+1),Cn)

∀j with −m ≤ j < m,

b ∈ H1(R,Cn), ψ(ξ, j−)−ψ(ξ, j+) = A∗j (ξ)b(ξ)∀ ξ and 0 < |j| ≤m}

(see Remark 2.1 for the notation). Furthermore, T ∗(ψ,b) = 0 if, and only if,L∗b = 0.

Proof. The domain D(T ∗) of the adjoint operator T ∗ is given by

D(T ∗) = {(ψ,b) ∈ L2(R, Y );

∃(ψ∗, b∗) ∈ L2(R, Y ) : 〈T (φ,a), (ψ,b)〉 = 〈(φ,a), (ψ∗, b∗)〉∀(φ,a) ∈ D(T )},

in which case T ∗(ψ,b) := (ψ∗, b∗). Thus, we consider the equation

(2.5)∞∫−∞

m∫−m

(dφdξ

− dφdη

)·ψdηdξ

+∞∫−∞

dadξ

−A0(ξ)a−∑j≠0

Aj(ξ)φ(ξ, j)

· b dξ

=∞∫−∞

m∫−mφ ·ψ∗ dηdξ +

∞∫−∞a · b∗ dξ.


Upon setting a = 0, so that φ(ξ,0) = 0, we obtain

(2.6)∞∫−∞

m∫−m

(dφdξ

− dφdη

)·ψdηdξ −

∞∫−∞

∑j≠0

Aj(ξ)φ(ξ, j) · b dξ

=∞∫−∞

m∫−mφ ·ψ∗ dηdξ.

If we restrict to test functions φ with φ(ξ, j) = 0 for all integers j, we see thatψ∗ = (∂η−∂ξ)ψ in L2(R× (j, j+1),Cn) for all j with −m ≤ j < m. Note thatthis definesψ∗ uniquely in L2(R×[−m,m],Cn). Using the notation introducedin Remark 2.1 and considering arbitrary test functions φ with φ(ξ,0) = 0, weobtain

∞∫−∞

m∫−m

(dφdξ

− dφdη

)·ψdηdξ

=∞∫−∞

m∫−mφ ·ψ∗ dηdξ +

m∫−m

∑j≠0

φ(ξ, j) · (ψ(ξ, j−)−ψ(ξ, j+)) dη,

and conclude that (2.6) is met for all φ with φ(ξ,0) = 0 provided

ψ(·, j−)−ψ(·, j+) = A∗j (·)b(·)

in L2(R,Cn) for all j ≠ 0. We return to (2.5) which, based on the results estab-lished above, reduces to

∫m−mφ(ξ,0) · (ψ(ξ,0−)−ψ(ξ,0+))dη

+∫∞−∞

(dadξ

−A0(ξ)a)· b dξ =

∫∞−∞a · b∗ dξ

and therefore to∫∞−∞

dadξ

· b dξ =∫∞−∞a · (b∗ +A∗0 (ξ)b −ψ(ξ,0−)+ψ(ξ,0+))dξ

for all a ∈ H1(R,Cn), since φ(·,0) = a for every (φ,a) ∈ D(T ). This showsthat b ∈ H1(R,Cn) and

b∗ = −dbdξ

−A∗0 b +ψ(·,0−)−ψ(·,0+).


Lastly, the proof that the null spaces of T ∗ and L∗ are isomorphic is quite anal-ogous to the proof of Lemma 2.2. The following argument is the key. Supposethat (ψ,b) ∈ D(T ∗) with T ∗(ψ,b) = 0. Using that (∂ξ − ∂η)ψ = 0 and thatψ(ξ, j−) = ψ(ξ, j+) + A∗j (ξ)b(ξ) for j > 0 with ψ(ξ,m−) = A∗m(ξ)b(ξ), itfollows inductively that

ψ(ξ, j−) =m∑k=jA∗k (ξ − k+ j)b(ξ − k+ j)

for j > 0. Hence,

ψ(ξ,0+) = ψ(ξ−1, 1−) =m∑j=0

A∗j (ξ − j)b(ξ − j).

Similarly, we obtain

ψ(ξ,0−) = ψ(ξ+1, −1+) = −−m∑j=0

A∗j (ξ − j)b(ξ − j)

and therefore

dbdξ

= −A∗0 (ξ)b +ψ(·,0−)−ψ(·,0+) = −A∗0 (ξ)b −m∑

j=−mA∗j (ξ − j)b(ξ − j).

We omit the remaining details. ❐As a consequence, T is Fredholm whenever L is.

Lemma 2.4. If L is Fredholm with index i, then T is Fredholm with the sameindex i.

Proof. On account of Lemma 2.2 and Lemma 2.3, we have dimN(L) =dim N(T ) and dim N(L∗) = dim N(T ∗). To show that the range of T is closed,assume that T (φn,αn) → (ψ,β) is a convergent sequence in R(T ). Definevn(ξ) := φn(ξ)(0) = αn(ξ) which gives a sequence (vn)n∈N in H1(R,Cn).The sequence (Lvn) converges in L2(R,Cn) to β. Since R(L) is closed, we knowthat Lvn → Lv∞ for some v∞ ∈ H1(R,Cn). Lemma 2.2 implies that (ψ,β) =T (φ∞, α∞) for φ∞(ξ) := v∞|[ξ−m,ξ+m] and α∞ := v∞. Thus, R(T ) is closed,and we conclude that R(T ∗) is also closed [8, Theorem IV.5.13]. In addition, wethen have codimR(L) = codimR(T ) since

codimR(L) = dim N(L∗) = dim N(T ∗) = codimR(T )

by the arguments at the beginning of the proof. ❐


3. THE CONSTANT-COEFFICIENT OPERATOR A0

For given matrices Aj with |j| ≤m, consider the constant-coefficient operator

(3.1) A0 : Y -→ Y ,(φa

)7 -→

dφdη

A0a+∑

1≤|j|≤mAjφ(j)

which is densely defined with domain D(A0) = Y 1 (see (2.3)). The characteristicequation associated with A0 is det∆(µ) = 0 where

∆(µ) :=m∑

j=−mAjejµ − µ id .

We are interested in proving the existence of exponential dichotomies for the equa-tion

dVdξ

=A0V,

since this equation with constant coefficients serves as a reference equation for themore general case that we consider in Section 4 below. To prove the existenceof exponential dichotomies for equations with constant coefficients, we first showthat A0 has only point spectrum and provide estimates for its resolvent. After-wards, we establish in Section 3.2 the completeness of eigenfunctions and proceedthen in Section 3.3 with the construction of exponential dichotomies in a fashionsimilar to Rustichini [15] who proved the existence of dichotomies for slightly differentoperators in C0([−m,m],Cn).

3.1. Spectrum, and resolvent estimates. We begin by establishing that thespectrum of A0 consists entirely of eigenvalues.

Lemma 3.1. The operatorA0 has only point spectrum. Moreover, µ ∈ C is in thespectrum of A0 if, and only if, det∆(µ) = 0. If µ is an eigenvalue, the correspondingeigenfunction is of the form (φ,a) with a ∈ Cn and φ(η) = aeηµ.

Proof. To determine the spectrum and the resolvent of A0 we have to discussthe equation

(A0 − µ)(φa

)=(ψb

)where ψ ∈ L2([−m,m],Cn) and b ∈ Cn. This equation is equivalent to

dφdη

− µφ = ψ

(A0 − µ)a+∑

1≤|j|≤mAjφ(j) = b.


Solving the first equation by the variations-of-constants formula shows that

φ(η) = eηµφ(0)+∫ η

0e(η−σ)µψ(σ)dσ.

Substituting this expression for φ into the second equation and exploiting thatφ(0) = a, we get

∆(µ)a = b −∑j≠0

Aj∫ j

0e(j−σ)µψ(σ)dσ.

Therefore, for any µ with det∆(µ) ≠ 0, i.e., for any µ that is not in the pointspectrum of A0, the resolvent of A0 is given by

φ(η) = eηµ∆(µ)−1

b −∑j≠0

Aj∫ j

0e(j−σ)µψ(σ)dσ

(3.2)

+∫ η

0e(η−σ)µψ(σ)dσ

a = ∆(µ)−1

b − ∑j≠0

Aj∫ j


. ❐

For later use, we derive estimates for the resolvent of A0. Rustichini [15]proved similar estimates for linear operators associated with constant-coefficientequations considered in C0([−m,m],Cn).

Lemma 3.2. Fix a constant κ > 0. There is then a constant M > 0 such that,for any µ with det∆(µ) ≠ 0 and |Reµ| > κ, we have

‖(A0 − µ)−1‖L(Y) ≤ M(

em|Reµ|

|Reµ| + ‖∆(µ)−1‖e2m|Reµ|

|Reµ|

).

Proof. We have shown that the equation

(A0 − µ)(φa

)=(ψb

)∈ Y

has the solution

φ(η) = eηµ∆(µ)−1

b − ∑j≠0

Aj∫ j


+ ∫ η0

e(η−σ)µψ(σ)dσ

a = ∆(µ)−1

b − ∑j≠0

Aj∫ j



whenever det∆(µ) ≠ 0 with ∆(µ) = ∑mj=−mAjejµ − µ id. Note that

∣∣∣∣∫ y0

e2(y−σ)Reµ dσ∣∣∣∣1/2

≤ e|yReµ|√2|Reµ| .

Using this inequality in combination with Holder’s inequality, we find that

|a| ≤ ‖∆(µ)−1‖|b| + ∑

j≠0

‖Aj‖e|jReµ|√2|Reµ|‖ψ‖L2

≤ ‖∆(µ)−1‖

(|b| +M1

em|Reµ|√2|Reµ|‖ψ‖L2

)

≤ ‖∆(µ)−1‖(

1+M1em|Reµ|√2|Reµ|

)∥∥∥∥∥(ψb

)∥∥∥∥∥Y

whereM1 := 2m max

−m≤j≤m‖Aj‖.

The same estimate appears when we bound the L2-norm of φ:

‖φ‖L2 ≤ ‖eηµ‖L2‖∆(µ)−1‖(

1+M1em|Reµ|√2|Reµ|

)∥∥∥∥∥(ψb

)∥∥∥∥∥Y+∥∥∥∥∥ e|ηReµ|√

2|Reµ|

∥∥∥∥∥L2

‖ψ‖L2

≤ em|Reµ|√2|Reµ|‖∆(µ)−1‖

(1+M1

em|Reµ|√2|Reµ|

)∥∥∥∥∥(ψb

)∥∥∥∥∥Y+ em|Reµ|

|Reµ| ‖ψ‖L2 .

For |Reµ| > κ, this implies the desired inequality. ❐

Using results of Bellman and Cooke [1], Rustichini demonstrated the follow-ing statement on the location of zeros of ∆(µ).

Lemma 3.3 ([15, Lemma 3.2]). Assume that det(A−mAm) ≠ 0, then thefollowing is true.

(i) For any κ > 0, there exists a constant M(κ) such that | Imµ| ≤ M(κ) for anyzero µ of ∆(µ) with |Reµ| ≤ κ. In particular, if there are no zeros of ∆(µ)on the imaginary axis, then there is a strip {µ; |Reµ| ≤ κ} that contains noeigenvalues of A0.

(ii) There exists a positive constant κ such that all zeros of ∆ are contained in theunion W+ ∪W− where

W± :={µ ∈ C; |Reµ| > κ,

∣∣∣∣Re(µ ± 1m

logµ)∣∣∣∣ ≤ κ} .


(iii) There exists a constant C > 0 such that

‖∆(µ)−1‖ ≤ Ce−m|Reµ|

along the curves |Re(µ ± (1/m) logµ)| = κ, that is, along curves where|µ|e±m|Reµ| is constant.

3.2. Completeness. An important property, which is not at all obvious fornon-selfadjoint operators such as A0, is completeness, i.e., the property that theclosure of the linear space spanned by the generalized eigenfunctions of A0 is theentire underlying function space Y . We demonstrate thatA0 is complete providedthat the matrices A−m and Am are not singular.

Theorem 3.1. Consider the operator A0 defined in (3.1). Assume that A0 hasno spectrum on the imaginary axis and that det(A−mAm) 6= 0. Let Eu be the closurein Y of the sum of generalized eigenspaces to all eigenvalues with positive real part.Similarly, let Es be the closure of the sum of the generalized eigenspace correspondingto eigenvalues with negative real part. We then have Eu ⊕ Es = Y .

Remark 3.1. An analogous statement holds if the spectrum has a center part,i.e., in the situation that there are eigenvalues on the imaginary axis.

Our proof of Theorem 3.1 is based on a characterization of completenessgiven by Verduyn Lunel in [19] (see Lemma 3.8 below). We begin by recallingsome facts from complex analysis.

Definition 3.1. Let X be a complex Banach space. An entire function F :C→ X is said to be of exponential type E(F) if

lim supr→∞

1r

log max0≤θ≤2π

‖F(reiθ)‖ =: E(F) <∞

exists.

In the following proposition, we summarize some properties of entire func-tions of exponential type.

Lemma 3.4. Let F1, F2 : C → C be entire functions of exponential type whichare polynomially bounded in the closed right half-plane Reµ ≥ 0. We then haveE(F1F2) = E(F1)+E(F2). If the quotient F1/F2 is an entire function, then E(F1/F2) =E(F1)− E(F2).

Note that the same holds for entire functions which are bounded in the lefthalf-plane. This symmetry will allow us later to relax some conditions. The fol-lowing lemma will prove useful below.

Lemma 3.5. If F : C → C is an entire function of exponential type that satisfies

lim supr→∞

1r

log |F(±ir)| = 0 and |F(µ)| ≤ M for µ ∈ R,


then F is constant.

Proof. The assertion is a consequence of [3, Thm. 6.2.4], which is a theo-rem by Duffin and Schaeffer [5], applied separately to F(µ) and F(−µ) with µrestricted to the upper half-plane in conjunction with Liouville’s Theorem. ❐

Using the identity

∆(µ)−1 = 1det∆(µ) cof ∆(µ),

where cof ∆(µ) is the matrix of cofactors, we can rewrite the solution of the equa-tion

(A0 − µ)(φa

)=(ψb

)

given in (3.2) as

φ(η) = eηµ

det∆(µ) cof ∆(µ)b − ∑

j≠0

Aj∫ j


(3.3)

+∫ η

0e(η−σ)µψ(σ)dσ

a = cof ∆(µ)det∆(µ)

b − ∑j≠0

Aj∫ j


.It is not hard to see that some of the functions involved in the above expressionare entire functions of exponential type.

Lemma 3.6. The exponential type of det∆(µ) ismn if, and only if, detA−m ≠0 or detAm ≠ 0. In this case the exponential type of each entry of the cofactor matrixcof ∆(µ) is m(n− 1).

Moreover, if detA−m ≠ 0, then e−mnµ det∆(µ) is of exponential type 2mn andbounded in the right half-plane, while the entries of e−m(n−1)µ cof ∆(µ) are boundedin the right half-plane and have exponential type 2m(n− 1).

Similarly, if detAm ≠ 0, then emnµ det∆(µ) is of exponential type 2mn andbounded in the left half-plane, while the entries of em(n−1)µ cof ∆(µ) are bounded inthe left half-plane and have exponential type 2m(n− 1).

Proof. The term det∆(µ) is a linear combination of terms of the form µqejµwhere 0 ≤ q ≤ n and −mn ≤ j ≤ mn. This implies that the exponential typeof det∆(µ) cannot be greater than mn. The coefficient of emnµ is detAm, whilethe coefficient of e−mnµ is detA−m. If at least one of those coefficients is nonzero,then the definition of exponential type with µ restricted to the real line shows thatE(det∆) = mn. On the other hand, the exponential type is strictly less than


mn if both detA−m and detAm vanish. Similar arguments apply to the entries ofcof ∆(µ). Note that, if detA±m ≠ 0, then each sub-determinant of A±m does notvanish either. The remaining estimates can be obtained by completely analogousarguments. ❐

In view of Lemma 3.5 it is also important to control the behavior of theresolvent along the imaginary axis.

Lemma 3.7. We have the following asymptotic behavior on the imaginary axis:

limr→∞

det∆(±ir)rn

= 1

limr→∞

(cof ∆(±ir))kkrn−1 = 1 for k = 1,2, . . . , n

limr→∞

(cof ∆(±ir))klrn−2 = (−1)k+l for k ≠ l.

Proof. These relations are simple consequences of the fact that along theimaginary axis the polynomial terms dominate the exponential terms. ❐

We use the following characterization of non-completeness.

Lemma 3.8 ([19, Lemma 3.2]). If B : X → X is an unbounded operator withmeromorphic resolvent, then the system of eigenfunctions and generalized eigenfunc-tions is not complete if, and only if, there exists a y∗ ∈ X∗ with y∗ ≠ 0 such that thefunction

µ 7 -→ 〈y∗, (B − µ)−1x〉is entire for every fixed x ∈ X.

We will show that no such y∗ exists in our situation. The explicit form ofthe resolvent (A0 − µ)−1 shows that it is indeed a meromorphic function so thatLemma 3.8 applies to A0.

Proof of Theorem 3.1. Assume that the system of eigenvectors and generalizedeigenvectors of A0 is not complete. Applying Lemma 3.8 and the Riesz represen-tation theorem to (3.3), we see that there are (φ,a) ∈ Y such that, for every fixed(ψ,b) ∈ Y , the function⟨(φa

), (A0 − µ)−1

(ψb

)⟩

= a · 1det∆(µ) cof ∆(µ)

b − ∑j≠0

Aj∫ j


+∫m−mφ(η) · 1

det∆(µ) cof ∆(µ)eηµb −

∑j≠0

Aj∫ j

0e(η+j−σ)µψ(σ)dσ

dη +


+∫m−mφ(η) ·

(∫ η0

e(η−σ)µψ(σ)dσ)

dη

is entire. In particular, this is true for ψ = 0 for which the above expressionreduces to⟨(

φa

), (A0 − µ)−1

(0b

)⟩= a · 1

det∆(µ) cof ∆(µ)b+∫m−mφ(η) · 1

det∆(µ) cof ∆(µ)eηµb dη.

If the right-hand side defines an entire function for all b ∈ Cn, then each compo-nent Fk(µ) of

F(µ) := a · cof ∆(µ)det∆(µ) +

∫m−mφ(η) · cof ∆(µ)

det∆(µ)eηµ dη

is an entire function. We prove that this implies that φ = 0 which leads to acontradiction.

We first show that each Fk(µ) satisfies the assumptions of Lemma 3.5 withc = 0. Note that, since detA−m ≠ 0, by Lemma 3.6 both the numerator andthe denominator of e−mnµFk(µ) are entire functions of exponential type 2mnwhich are bounded in the right half-plane. Assuming that Fk is itself entire, wecan conclude from Lemma 3.4 that Fk is of exponential type 0. Regarding thebehavior of Fk along the imaginary axis, Lemma 3.7 shows that Fk(ir) convergesto 0 for r → ±∞. This implies directly the weaker statement

lim supr→∞

1r

log |Fk(±ir)| ≤ 0

which is used in Lemma 3.5. The last hypothesis that needs to be checked is theboundedness on the real axis. Since by assumption F is an entire function, weonly need to be concerned about the behavior for µ → ±∞. Since detA−m ≠ 0and detAm ≠ 0, we conclude that

|det∆(µ)| ≥ Cemn|µ|

for some constant C and |µ| sufficiently large. On the other hand, cof ∆(µ) is ofexponential type m(n− 1), whence we can estimate∣∣∣∣∫m−mφ(η) · cof ∆(µ)eηµb dη

∣∣∣∣ ≤ ∣∣∣∣∫m−mφ(η)em(n−1)|µ|em|µ|b dη∣∣∣∣

which implies that Fk(µ) is uniformly bounded for large |µ|.


Hence, Lemma 3.5 implies that each Fk is constant. Using the behavior alongthe imaginary axis, we conclude immediately that this constant is zero so that

Fk(µ)det∆(µ) = a · cof ∆(µ)+ ∫m−mφ(η) · cof ∆(µ)eηµ dη = 0.

This implies that

−a =∫m−mφ(η)eηµ dη =

∫ 2m

0φ(η−m)emµeηµ dη

is a constant independent of µ. In other words, if we extend φ to R by settingφ(η) = 0 for |η| > m, then the constant function −a would be the Laplacetransform of the function φ(η−m)emµ which is in L2 and has compact support.However, as the Laplace of a function with compact support, −a would have tobe integrable along each line Reµ = const. This would imply a = 0, and byinverse Laplace transform φ = 0. Therefore, by Lemma 3.8, the operator A0 hasa complete system of eigenvectors and generalized eigenvectors. ❐

3.3. Exponential dichotomies. We are now in a position to establish theexistence of exponential dichotomies for the equation

(3.4)dVdξ

=A0V

with constant coefficients. We assume that A0 is hyperbolic, i.e., that ∆(µ) ≠ 0for all µ ∈ iR. Recall from Lemma 3.3 that the distance from the spectrum of A0

to the imaginary axis is strictly positive. Let Es and Eu be the closure in Y of thegeneralized eigenspaces associated with all eigenvalues of A0 that have negativeand positive real part, respectively.

Proposition 3.1. Assume that det(A−mAm) ≠ 0, and choose κ > 0 such thatκ is smaller than the distance from the spectrum of A0 to the imaginary axis. Thereexists then a strongly continuous semigroup Φs(ξ) : Es → Es defined for ξ ≥ 0 and aconstant K such that Φs(0) = id and

‖Φs(ξ)‖L(Y) ≤ Ke−κξ.

For any V0 ∈ Es ∩ Y 1, the function V(ξ) = Φs(ξ)V0 is differentiable for ξ > 0and satisfies (3.4) with V(0) = V0. Analogous statements hold for Φu(ξ) : Eu → Eu

defined for ξ ≤ 0.

Note that, for V0 ∈ Es, the function V(ξ) = Φs(ξ)V0 is a mild solution of(3.4), i.e., it satisfies the integral equation

V(ξ) = V0 +∫ ξ

0A0V(ζ)dζ

for all ξ ≥ 0.


Proof. It suffices to prove the statement for Φs(ξ). We begin by constructingsemigroups on the subspace Es which is defined as the sum of all generalizedeigenspaces of eigenvalues µ of A0 with Reµ < 0.

We construct the semigroup Φs via an integral representation. We begin bychoosing curves Γ1 and Γ2 in C by

Γ1 :={µ ∈ C : Re

(µ + 1

mlogµ

)= κ, Imµ < 0, Reµ ≤ −κ

}Γ2 :=

{µ ∈ C : Re

(µ + 1

mlogµ

)= κ, Imµ > 0, Reµ ≤ −κ

}

where κ has been defined in Lemma 3.3. Note that Γ1 and Γ2 can be parametrizedby µ = x+ iy(x) where y ′(x) = O(e−mx) as x → −∞ (see [15, p. 140]). Lastly,the curve Γ3 joins Γ1 and Γ2 along the line Reµ = −κ. By Lemma 3.3, the curveΓ := Γ1 ∪ Γ2 ∪ Γ3 lies to the right of the negative part of the spectrum.

For V0 ∈ Es and ξ0 > 2m, we have∥∥∥∥∥ 12π i

∫Γ1 eµξ0(A0 − µ)−1V0 dµ

∥∥∥∥∥Y

≤ 12π

∫Γ1 eξ0 Reµ‖(A0 − µ)−1‖dµ ‖V0‖Y

≤ 12π

∫Γ1 eξ0 ReµM

(em|Reµ|

|Reµ| + Ce−m|Reµ| e2m|Reµ|

|Reµ|

)dµ ‖V0‖Y

≤ 12π

∫ 0

−∞eξ0xe−mxM

1+ C|x| e−mx dx ‖V0‖Y

= K(ξ0)‖V0‖Y

where we used the above parametrization of Γ1 to evaluate the line integral. Notethat the last integral converges for ξ0 > 2m. Since the same calculation applies tothe integral along Γ2, we can define an operator Φs(ξ) on Es for ξ ≥ ξ0 > 2m by

(3.5) Φs(ξ)V0 := 12π i

∫Γ eµξ(A0 − µ)−1V0 dµ.

In particular, for ξ ≥ ξ0 > 2m, we have

(3.6) ‖Φs(ξ)V0‖Y ≤∥∥∥∥ 1

2π i

∫Γ eµξ(A0 − µ)−1V0 dµ

∥∥∥∥ ≤ e−κξK(ξ0)eκξ0‖V0‖Y

for some K(ξ0) ≥ 1, which gives exponential decay for ξ ≥ ξ0 > 2m. Note that,for V0 ∈ Es ∩ Y 1, the function V(ξ) = Φs(ξ)V0 satisfies (3.4) for ξ > 2m withV(0) = V0 (see [13]).


Next, we define a filtration of Es by finite-dimensional generalized eigenspacesEsj ⊂ Y 1 with j ∈ N (see [15]). For any V0 ∈ Es

j , we can readily solve (3.4)on R+ and get a solution V(ξ) that is differentiable with values in Y . On theinterval [ξ0,∞), the solution V(ξ) coincides with Φs(ξ)V0 as defined in (3.5). Inparticular, with V0 = (φ,φ(0)) and V(ξ) = (vξ, v(ξ)), the function v(ξ) is asolution of

(3.7) v′(ξ) =m∑

j=−mAjv(ξ + j)

for ξ ≥ 0 with v|[−m,m] = φ. Hence, using (3.6), we obtain the L2-estimate

(3.8) ‖v‖L2([m+ε,3m+ε]) ≤ K(2m+ ε)‖φ‖L2

for any 0 < ε < 1/2. Our goal is to estimate the solution v in L2([m,m+ ε]) interms of the L2-norm of the initial condition φ. Let C1 be a bound for the normsof the matrices Aj . Using that v satisfies (3.7), we then get the H1-estimate

‖v‖H1([m+ε,m+2ε]) ≤mC1(1+K(2m+ ε))‖φ‖L2 ,

and finally

(3.9) |v(m+ ε)| ≤ C2(1+K(2m+ ε))‖φ‖L2

by Sobolev’s embedding theorem for some C2 that may depend on ε. Multiplying(3.7) by v(ξ), we obtain

v′(ξ) · v(ξ) = A0v(ξ) · v(ξ)+∑j≠0

Ajv(ξ + j) · v(ξ)

and therefore

12

ddξ|v(ξ)|2 ≤ C1|v(ξ)|2+

∑j≠0

C1|v(ξ+j)| |v(ξ)| ≤ C3|v(ξ)|2+∑j≠0

C1|v(ξ+j)|2.

Integrating this inequality over [ξ,m+ ε] with ξ ∈ [m,m+ ε] and ε ≤ 1/2, weobtain

|v(ξ)|2 ≤ |v(m+ ε)|2 +∫m+εξ

C3|v(ζ)|2 +∑j≠0

C1|v(ζ + j)|2 dζ

≤ |v(m+ ε)|2 +∫m+εξ

C3|v(ζ)|2 +∑j<0

C1|φ(ζ + j)|2 dζ + C4‖φ‖2

L2

≤ C(ε)‖φ‖2 +∫m+εξ

C3|v(ζ)|2 dζ


where we used that v(ξ + j) = φ(ξ + j) for j < 0 and exploited the estimates(3.8) and (3.9). Using Gronwall’s inequality, we get

(3.10) ‖v‖2L2([m,m+ε]) ≤ C‖φ‖2

Y .

In summary, from (3.6) and (3.10), we finally conclude that

(3.11) ‖vξ‖L2([−m,m]) ≤ K‖φ‖Y

for all ξ ≥ 0, where v(ξ) is the solution of (3.7) with initial conditionφ associatedwith a given V0 = (φ,φ(0)) ∈ Es

j . Note that the constant K that appears in (3.11)does not depend on j (and not on V0). In addition, we have

(3.12) ‖vξ‖H1([−m,m]) ≤ CKe−κξ‖φ‖Y

for ξ ≥m by using (3.7) and (3.6).Having established these uniform estimates on Es

j , it remains to extend thesemigroup Φs(ξ) from Es

j to the closure Es, while maintaining the estimates (3.6).This can be done in a straightforward manner by approximating initial conditionsin Es by elements in Es

j in the L2-sense and using compactness properties impliedby theH1-estimate (3.12) on any given bounded interval in ξ. We omit the detailsas they are similar (and in fact easier) than those given in [15]. ❐

4. FREDHOLM PROPERTIES OF T IMPLY THE EXISTENCE OFDICHOTOMIES

In this section, we prove that the non-autonomous equation (2.1) has an expo-nential dichotomy. The arguments are similar to those used in [18] in the caseof modulated travelling waves. For this reason, we give an outline of the proofand provide details only where the arguments for forward-backward delay equa-tions are different. The main strategy is to compare the non-autonomous operatorwith the constant-coefficient operator for which the existence of exponential di-chotomies has been shown in the previous section.

After extending the operator T in Section 4.1 to a larger function space, weprove Theorem 1.1 in Sections 4.2 and 4.3, and Theorem 1.2 in Section 4.4.

4.1. The extension S of T . We consider the operator

T : L2(R, Y ) -→ L2(R, Y ), V 7 -→ dVdξ

−AV

with domain D(T ) given in (2.4). The adjoint T ∗ of T is also densely defined inL2(R, Y ) with domain D(T ∗) given in Lemma 2.3. Alternatively, we can considerT ∗ as a bounded operator, denoted by T ∗, from D(T ∗) to L2(R, Y ). Here, we


consider D(T ∗) as a Banach space equipped with the graph norm. We denote by

S the adjoint operator(T ∗

)∗of T ∗, so that

S : L2(R, Y ) -→ D(T ∗)∗.

Note that S restricted to D(T ) coincides with T . We remark that the notation(T ∗)ad instead of S is used in [18].

By definition, the equation SU = G means that (T ∗W,U) = (W,G) for allW ∈ D(T ∗). The brackets denote the duality pairing of D(T ∗) and D(T ∗)∗.In other words, SU = G is a shortcut for

−∫∞−∞〈∂ξW +A(ξ)∗W,U〉Y dξ =

∫∞−∞〈W,G〉Y dξ, ∀W ∈ D(T ∗)

where the scalar products 〈·, ·〉Y are interpreted in the sense of distributions.

Lemma 4.1. Assume that T is Fredholm, then S is also Fredholm with the sameindex. Furthermore, N(S) = N(T ).

Proof. The statements are consequences of [8, Section III §5.5 on p. 168]and [8, Section IV §5.3 on p. 236]. ❐

4.2. T is invertible. It is convenient to consider first the case that T is in-vertible before proceeding to the more general case that T is only Fredholm.

Lemma 4.2 ([18, Lemma 5.2]). Assume that T is invertible. For any ξ0 ∈R and any G0 ∈ Y , define G(ξ,η) := G0(η)δ(ξ − ξ0) where δ denotes the δ-distribution. There is then a unique solution U ∈ L2(R, Y ) of the equation

SU = G.

The restrictions of U to (−∞, ξ0] and [ξ0,+∞) belong to C0((−∞, ξ0], Y ) andC0([ξ0,+∞), Y ), respectively. The limits U+(ξ0) := limξ↘ξ0 U(ξ) and U−(ξ0) :=limξ↗ξ0 U(ξ) exist and satisfy the jump condition U+(ξ0)− U−(ξ0) = G0. In addi-tion, we have the estimate

‖U‖L∞(R,Y ) + ‖U‖L2(R,Y ) ≤ C|G0|Ywhere the constant C is independent of G0.

Proof. Note that the equation SU = G is well-defined since G ∈ D(T )∗ byLemma 2.1. Without loss of generality, let ξ0 = 0.

The proof is based on the comparison with an appropriate reference equation.Choose matrices Aref

j for 0 ≤ |j| ≤m such that det(Aref−mArefm ) ≠ 0 and such that

det∆(µ) ≠ 0 for all µ ∈ iR where

∆(µ) = m∑j=−m

Arefj ejµ − µ id .


Define

Aref :

(φa

)7 -→

dφdη

Aref0 a+

∑1≤|j|≤m

Arefj φ(j)

so that A0 is hyperbolic. We write T as

T = Tref +B

where Tref = ddξ −Aref and

B(ξ) =Aref −A(ξ) :

(φa

)7 -→

0

(Aref0 −A0(ξ))a+

∑1≤|j|≤m

(Arefj −Aj(ξ))φ(j)

.The strategy is to first seek a solution V ∈ L2(R, Y ) of the reference equationSrefV = G and afterwards a continuous solution U that satisfies SU = −BV . Thesum U := V + U then satisfies

SU = S(V + U) = SrefV +BV + SU = G

as desired.We begin by solving the equation

SrefV =(

ddξ−Aref

)V = G.

According to Proposition 3.1, the equation

(4.1)dVdξ

= ArefV

has exponential dichotomies. Let Pref and (id−Pref) be the corresponding pro-jections such that (4.1) generates exponentially decaying C0-semigroups on bothR(Pref) and R(id−Pref). For any G0 ∈ Y 1, define

V(ξ) :={

eArefPrefξPrefG0 for ξ > 0−eAref(id−Pref)ξ(id−Pref)G0 for ξ < 0.

By standard semigroup theory, the function V is differentiable for ξ ≠ 0 andsatisfies (4.1) for ξ ≠ 0. Furthermore, the limits V 0+ := limξ↘0 V(ξ) and V 0− :=


limξ↗0 V(ξ) exist. For any test function χ ∈ C∞0 (R, Y 1), we obtain

−∫∞−∞

⟨V,

dχdξ

+A∗refχ

⟩Y

dξ

= −∫∞

0

⟨eArefPrefξPrefG0,

dχdξ

+A∗refχ

⟩Y

dξ

+∫ 0

−∞

⟨eAref(id−Pref)ξPrefG0,

dχdξ

+A∗refχ

⟩Y

dξ

=⟨eArefPrefξPrefG0, χ

⟩Y

∣∣∣ξ=0

+⟨eAref(id−Pref )ξPrefG0, χ

⟩Y

∣∣∣ξ=0

= V 0+ − V 0

− = 〈G0, χ(0)〉Y =∫∞−∞〈G0δ(ξ), χ〉Y dξ

using integration by parts and the fact that V satisfies (4.1) for ξ ≠ 0. This showsthat, for G0 ∈ Y 1, V satisfies SrefV = G. For G0 ∈ Y , we define a solutionV ∈ L2(R, Y ) of SrefV = G by approximating G0 by a sequence in Y 1 and usingthe strong continuity of the semigroup (see [18, Section 5.3.1] for details).

In the next step, we solve

(4.2) SU = −BVthat involves the solution V of the reference equation. The right-hand side −BVof (4.2) is in L2(R, Y ) since its first component vanishes completely, while thesecond component is continuous except at ξ = 0 and decays exponentially asξ → ±∞. Since S restricted to D(T ) coincides with T , and since both operatorsare invertible by assumption and by Lemma 4.1, we can solve (4.2) and obtain aunique solution U = (φ, a) ∈ D(T ). Using the fact that the first component of−BV vanishes, we see that φ(ξ, η) = a(ξ+η) with a ∈ H1(R,Cn). In particular,U is continuous on R with values in Y and

‖U‖H1(R,Y ) + ‖U‖L2(R,Y 1) ≤ C‖V‖L2(R,Y ).

Lastly, as mentioned earlier, the sum U = U+V satisfies all the properties stated inthe lemma. In particular, it is continuous on R− and R+, and the jump at ξ = 0is exactly the jump of V at ξ = 0. Therefore, U+(0)−U−(0) = G0. ❐

The previous lemma allows us to define a continuous, injective map

Π(ξ0) : Y -→ Y × Y , G0 7 -→ (U+(ξ0),U−(ξ0)).

Using the canonical projections Pi(ξ0) : Y × Y → Y defined by Pi(ξ0)(U1, U2) =Ui for i = 1,2, we have the relation

G0 = P1(ξ0)Π(ξ0)G0 − P2(ξ0)Π(ξ0)G0


for any G0 ∈ Y .

Lemma 4.3 ([18, Lemma 5.3]). Suppose that T is invertible. The imagesR(Pi(ξ0)Π(ξ0)) are then closed subspaces, and we have R(P1(ξ0)Π(ξ0)) ⊕R(P2(ξ0)Π(ξ0)) = Y for each ξ0 ∈ R.

We can then define a family of projections P(ξ) such that R(P(ξ)) =R(P1(ξ)Π(ξ)) and N(P(ξ)) = R(P2(ξ)Π(ξ)). Note that, for anyU+ ∈ R(P(ξ0)),there exists a unique strong solution V s(ξ) of (2.2), defined for ξ > ξ0, with initialvalue V s(ξ0) = U+. Analogously, for any U− ∈ N(P(ξ0)), there exists a uniquestrong solution Vu(ξ) of (2.2), defined for ξ < ξ0, with initial value Vu(ξ0) = U−.This projections define the desired exponential dichotomies.

Lemma 4.4 ([18, Lemma 5.5]). The family P(ξ) of projections together withthe solutions V s and Vu defines an exponential dichotomy on R.

4.3. T is Fredholm. Again, we closely follow the proof given in [18] forparabolic equations in cylinders. Since most of the details are identical, save fornotation, we comment only on those two parts of the proof where the argumentsin [18, Section 5.3.2] need to be modified. First, the proof given in [18, Sec-tion 5.3.2] uses the fact that the null spaces of T and S coincide. We proved inLemma 4.1 that this is always the case. Second, the following weak uniquenesscondition (U1) has been used in [18].(U1) If V is in the null space of T or the adjoint operator T ∗ and V(ξ0) = 0 forsome ξ0, then V vanishes identically.

This assumption is, however, equivalent to Hypothesis 1.1 on account ofLemma 2.2. Now that the above two properties are established, the abstractproof in [18] applies to our situation and gives the existence of exponential di-chotomies. This completes the proof of Theorem 1.1 on the existence of expo-nential dichotomies.

4.4. Proof of Theorem 1.2. It suffices to show that the hypotheses of Theo-rem 1.2 imply the hypotheses of Theorem 1.1 which we proved in the last section.Proposition 2.1 which is due to Mallet-Paret [11] implies that L is Fredholm pro-vided it is asymptotically hyperbolic. We established in Lemma 2.4 that T is Fred-holm whenever L is. It therefore remains to demonstrate that Hypothesis 1.1 is aconsequence of the following Hypothesis 4.1 which we assumed in Theorem 1.2.

Hypothesis 4.1. Assume that neither detAm(ξ) nor detA−m(ξ) vanish onany open interval.

To establish that Hypothesis 1.1 follows from Hypothesis 4.1, suppose thata ∈ N(L) such that a(ξ) vanishes identically on the interval [ξ0−m, ξ0+m].We want to prove that this implies that a(ξ) = 0 for all ξ ∈ R and argue bycontradiction. Note that, by Lemma 2.2, a is a classical solution of (2.1). Thus,without loss of generality, we assume that

ξ1 := inf{ξ ≥ ξ0 : a(ξ +m) ≠ 0}


exists and is finite. Since a(ξ + j) = 0 for |j| ≤m and ξ0 ≤ ξ ≤ ξ1, we have

a′(ξ) =m∑

j=−mAj(ξ)a(ξ + j) = Am(ξ)a(ξ +m)

for all ξ ∈ [ξ1, ξ1+1). By the definition of ξ1, there exists an open and non-empty interval J ⊂ (ξ1, ξ1 + 1) such that a(ξ +m) ≠ 0 for any ξ ∈ J. Thus, weconclude that

a′(ξ) = Am(ξ)a(ξ +m) ≠ 0

for some ξ ∈ J, since detAm(ξ) does not vanish identically on J by Hypothe-sis 4.1. This contradicts the definition of ξ1.

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[17] B. SANDSTEDE, Stability of travelling waves. In: “Handbook of Dynamical Systems II”(B. Fiedler, ed.). Elsevier, (2002) 983–1055.

[18] B. SANDSTEDE AND A. SCHEEL, On the structure of spectra of modulated travelling waves,Math. Nachr. 232 (2001), 39–93.

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JORG HARTERICH :Institut fur Mathematik IFreie Universitat BerlinArnimallee 2-614195 Berlin, Germany

BJORN SANDSTEDE :Department of MathematicsThe Ohio State University231 West 18th AvenueColumbus, OH 43210, U.S.A.

ARND SCHEEL :School of MathematicsUniversity of Minnesota206 Church Street SEMinneapolis, MN 55455, U.S.A.

Received : June 12th, 2001.

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